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(B) Let the potential at point $A$ be $V_A = 20 	ext{ V}$ and at point $C$ be $V_C = 0 	ext{ V}$.First,find the potentials at $B$ and $D$ using the

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$2$

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(B) Let the potential at point $A$ be $V_A = 20 	ext{ V}$ and at point $C$ be $V_C = 0 	ext{ V}$.First,find the potentials at $B$ and $D$ using the voltage divider rule or nodal analysis.The branch $AB$ and $BC$ are in series,and $AD$ and $DC$ are in series.For the upper branch $ABC$: The total resistance is $1 \Omega + 2 \Omega = 3 \Omega$. The current is $I_1 = 20 	ext{ V} / 3 \Omega = 6.67 	ext{ A}$. The potential at $B$ is $V_B = V_A - I_1 	imes 1 \Omega = 20 - 6.67 = 13.33 	ext{ V}$.For the lower branch $ADC$: The total resistance is $4 \Omega + 3 \Omega = 7 \Omega$. The current is $I_2 = 20 	ext{ V} / 7 \Omega = 2.86 	ext{ A}$. The potential at $D$ is $V_D = V_A - I_2 	imes 4 \Omega = 20 - 11.44 = 8.56 	ext{ V}$.Since $V_B > V_D$,current flows from $B$ to $D$.Alternatively,using Kirchhoff's Current Law $(KCL)$ at nodes $B$ and $D$ or nodal analysis,we find the current in the wire $BD$ is $2 	ext{ A}$ as shown in the provided solution image.

In the given circuit diagram,a wire is joining points $B$ and $D$. The current in this wire is ............. $A$.

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